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Physics / Chapter 2: Motion Along a Straight Line / Section 2-2

Position and Displacement — Question 3

Section 2-2 · Physics pages 156–157

Question 3 Signed changes cancel; traveled lengths accumulate A walker follows a straight east-positive path through the positions A : −6 m, B : 4 m, C : −1 m, D : 7 m, in the order A, B, C, D. Initially assume the walker never reverses within a leg between successive listed stops. x (m) −6 A −1 C 4 B 7 D first leg second leg third leg Travel order A to B to C to D. Vertical separation of the leg arrows is for clarity, not another spatial direction. Distance traveled is the accumulated length of the actual path, regardless of direction. For a leg with no reversal it equals |xf −xi|. (a) Find each signed leg displacement, the total displacement, and total distance traveled. (b) Show algebraically why adding signed leg displacements gives the endpoint displacement. Explain why applying an absolute value before adding gives a different result here. (c) Let R be the accumulated eastward length and L the accumulated westward length. Find both and explain why the excess of total distance over the positive net displacement equals 2L. (d) Now remove the no-reversal-within-a-leg assumption, keeping only the ordered stop observations. Which result remains exact, and what can be said about total distance? Give a longer compatible route.
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Question 3 - Solution Signed changes cancel; traveled lengths accumulate x (m) −6 A −1 C 4 B 7 D first leg second leg third leg (a) Keep the signs until the requested quantity is clear. The leg displacements are +10, −5, +8 m . Therefore ∆x = 10 −5 + 8 = +13 m , D = 10 + 5 + 8 = 23 m . The net change is eastward; the traveled length includes the westward leg too. (b) Internal endpoint coordinates cancel in a signed sum. (xB −xA) + (xC −xB) + (xD −xC) = xD −xA. (xB −xA) + (xC −xB) + (xD −xC) = xD −xA. The repeated intermediate coordinates cancel. Absolute values remove the opposite signs that permit this cancellation: they count lengths rather than net changes. Thus |∆x| = 13 m is not the same as the 23 m path length. (c) Opposing travel is counted twice in the excess. D −|∆x| = (R + L) −|R −L| = 2 min(R, L) = 10 m. Here R = 10 + 8 = 18 m and L = 5 m . In general, D = R + L, ∆x = R −L, D −∆x = 2L. For this eastward net change, D −|∆x| = 23 −13 = 10 m = 2L. Each metre westward adds path length and also offsets a metre of eastward progress. (d) Endpoints fix displacement even when the path is hidden. The total displacement remains +13 m. The sum 23 m becomes a lower bound on distance, since any continuous path between successive stops is at least their separation. For example, during the first leg the walker could go from −6 to 5 and then back to 4 m before completing the listed later legs. The distance becomes 11 + 1 + 5 + 8 = 25 m. Without an extra constraint there is no finite upper bound on compatible traveled length.

Question 3

Signed changes cancel; traveled lengths accumulate

A walker follows a straight east-positive path through the positions A:−6m,B:4m,C:−1m,D:7m,A:-6\ \mathrm{m},\quad B:4\ \mathrm{m},\quad C:-1\ \mathrm{m},\quad D:7\ \mathrm{m}, in the order A, B, C, D. Initially assume the walker never reverses within a leg between successive listed stops.

Diagram shown in the original SVG worksheet.

Distance traveled is the accumulated length of the actual path, regardless of direction. For a leg with no reversal it equals |xf−xi||x_f-x_i|.

  1. Find each signed leg displacement, the total displacement, and total distance traveled.

  2. Show algebraically why adding signed leg displacements gives the endpoint displacement. Explain why applying an absolute value before adding gives a different result here.

  3. Let RR be the accumulated eastward length and LL the accumulated westward length. Find both and explain why the excess of total distance over the positive net displacement equals 2L2L.

  4. Now remove the no-reversal-within-a-leg assumption, keeping only the ordered stop observations. Which result remains exact, and what can be said about total distance? Give a longer compatible route.

Question 3 - Solution

Signed changes cancel; traveled lengths accumulate

Diagram shown in the original SVG worksheet.

(a) Keep the signs until the requested quantity is clear.

The leg displacements are +10,−5,+8m\boxed{+10,-5,+8\ \mathrm{m}}. Therefore Δx=10−5+8=+13m,D=10+5+8=23m.\Delta x=10-5+8=\boxed{+13\ \mathrm{m}},\qquad D=10+5+8=\boxed{23\ \mathrm{m}}. The net change is eastward; the traveled length includes the westward leg too.

(b) Internal endpoint coordinates cancel in a signed sum.

(xB−xA)+(xC−xB)+(xD−xC)=xD−xA.(x_B-x_A)+(x_C-x_B)+(x_D-x_C)=x_D-x_A.

(xB−xA)+(xC−xB)+(xD−xC)=xD−xA.(x_B-x_A)+(x_C-x_B)+(x_D-x_C)=x_D-x_A. The repeated intermediate coordinates cancel. Absolute values remove the opposite signs that permit this cancellation: they count lengths rather than net changes. Thus |Δx|=13m|\Delta x|=13\ \mathrm{m} is not the same as the 23 m path length.

(c) Opposing travel is counted twice in the excess.

D−|Δx|=(R+L)−|R−L|=2min⁡(R,L)=10m.D-|\Delta x|=(R+L)-|R-L|=2\min(R,L)=10\ \mathrm{m}.

Here R=10+8=18mR=10+8=\boxed{18\ \mathrm{m}} and L=5mL=\boxed{5\ \mathrm{m}}. In general, D=R+L,Δx=R−L,D−Δx=2L.D=R+L,\quad\Delta x=R-L,\quad D-\Delta x=2L. For this eastward net change, D−|Δx|=23−13=10m=2LD-|\Delta x|=23-13=10\ \mathrm{m}=2L. Each metre westward adds path length and also offsets a metre of eastward progress.

(d) Endpoints fix displacement even when the path is hidden.

The total displacement remains +13m+13\ \mathrm{m}. The sum 23 m becomes a lower bound on distance, since any continuous path between successive stops is at least their separation. For example, during the first leg the walker could go from −6-6 to 55 and then back to 4m4\ \mathrm{m} before completing the listed later legs. The distance becomes 11+1+5+8=25m11+1+5+8=25\ \mathrm{m}. Without an extra constraint there is no finite upper bound on compatible traveled length.

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